Chern classes of tautological bundle over the Grassmannian G(2,4)
$begingroup$
I've the following problem:
I know how to calculate Chern classes of the tautological bundle over the Grassmannian $G=G(2,4)$ using the Schubert calculus. If I am right, the Chern character should be
$$1-sigma_1+sigma_{1,1}$$
where $sigma_{i_1,i_2}$ indicates the Schubert cycle corresponding to the Schubert variety $Sigma_{i_1,i_2}={Lambdain Gmiddim(V_{2-i_j+j}capLambda)geq j;forall j}$ (here ${V_j}subset V$ is a flag in the 4-dimensional vector space $V$). Now these classes are element in the Chow ring, but I want to work with classes in the integral cohomology group. How can I do this translation?
Thank you!
algebraic-geometry schubert-calculus
edited Jan 2 at 0:49
Matt Samuel
38.7k63769
asked Jul 15 '14 at 14:07
ClaCla
638411
$endgroup$
migrated from mathoverflow.net Jul 15 '14 at 22:07
This question came from our site for professional mathematicians.
add a comment |
$begingroup$
I've the following problem:
I know how to calculate Chern classes of the tautological bundle over the Grassmannian $G=G(2,4)$ using the Schubert calculus. If I am right, the Chern character should be
$$1-sigma_1+sigma_{1,1}$$
where $sigma_{i_1,i_2}$ indicates the Schubert cycle corresponding to the Schubert variety $Sigma_{i_1,i_2}={Lambdain Gmiddim(V_{2-i_j+j}capLambda)geq j;forall j}$ (here ${V_j}subset V$ is a flag in the 4-dimensional vector space $V$). Now these classes are element in the Chow ring, but I want to work with classes in the integral cohomology group. How can I do this translation?
Thank you!
algebraic-geometry schubert-calculus
edited Jan 2 at 0:49
Matt Samuel
38.7k63769
asked Jul 15 '14 at 14:07
ClaCla
638411
$endgroup$
migrated from mathoverflow.net Jul 15 '14 at 22:07
This question came from our site for professional mathematicians.
3
$begingroup$
The natural map $CH(G)rightarrow H^*(G,mathbb{Z})$ is a ring isomorphism.
$endgroup$
– abx
Jul 15 '14 at 14:28
$begingroup$
I know that a such map exists, but I don't how it works. So in my specific example, what is the image of $-sigma_1$? And that of $sigma_{1,1}$? I never did such a calculation..
$endgroup$
– Cla
Jul 17 '14 at 9:25
add a comment |
$begingroup$
I've the following problem:
I know how to calculate Chern classes of the tautological bundle over the Grassmannian $G=G(2,4)$ using the Schubert calculus. If I am right, the Chern character should be
$$1-sigma_1+sigma_{1,1}$$
where $sigma_{i_1,i_2}$ indicates the Schubert cycle corresponding to the Schubert variety $Sigma_{i_1,i_2}={Lambdain Gmiddim(V_{2-i_j+j}capLambda)geq j;forall j}$ (here ${V_j}subset V$ is a flag in the 4-dimensional vector space $V$). Now these classes are element in the Chow ring, but I want to work with classes in the integral cohomology group. How can I do this translation?
Thank you!
algebraic-geometry schubert-calculus
edited Jan 2 at 0:49
Matt Samuel
38.7k63769
asked Jul 15 '14 at 14:07
ClaCla
638411
$endgroup$
I've the following problem:
I know how to calculate Chern classes of the tautological bundle over the Grassmannian $G=G(2,4)$ using the Schubert calculus. If I am right, the Chern character should be
$$1-sigma_1+sigma_{1,1}$$
where $sigma_{i_1,i_2}$ indicates the Schubert cycle corresponding to the Schubert variety $Sigma_{i_1,i_2}={Lambdain Gmiddim(V_{2-i_j+j}capLambda)geq j;forall j}$ (here ${V_j}subset V$ is a flag in the 4-dimensional vector space $V$). Now these classes are element in the Chow ring, but I want to work with classes in the integral cohomology group. How can I do this translation?
Thank you!
algebraic-geometry schubert-calculus
algebraic-geometry schubert-calculus
edited Jan 2 at 0:49
Matt Samuel
38.7k63769
asked Jul 15 '14 at 14:07
ClaCla
638411
edited Jan 2 at 0:49
Matt Samuel
38.7k63769
asked Jul 15 '14 at 14:07
ClaCla
638411
edited Jan 2 at 0:49
Matt Samuel
38.7k63769
edited Jan 2 at 0:49
Matt Samuel
38.7k63769
edited Jan 2 at 0:49
Matt Samuel
38.7k63769
38.7k63769
asked Jul 15 '14 at 14:07
ClaCla
638411
asked Jul 15 '14 at 14:07
ClaCla
638411
asked Jul 15 '14 at 14:07
ClaCla
638411
638411
migrated from mathoverflow.net Jul 15 '14 at 22:07
This question came from our site for professional mathematicians.
migrated from mathoverflow.net Jul 15 '14 at 22:07
This question came from our site for professional mathematicians.
3
$begingroup$
The natural map $CH(G)rightarrow H^*(G,mathbb{Z})$ is a ring isomorphism.
$endgroup$
– abx
Jul 15 '14 at 14:28
$begingroup$
I know that a such map exists, but I don't how it works. So in my specific example, what is the image of $-sigma_1$? And that of $sigma_{1,1}$? I never did such a calculation..
$endgroup$
– Cla
Jul 17 '14 at 9:25
add a comment |
3
$begingroup$
The natural map $CH(G)rightarrow H^*(G,mathbb{Z})$ is a ring isomorphism.
$endgroup$
– abx
Jul 15 '14 at 14:28
$begingroup$
I know that a such map exists, but I don't how it works. So in my specific example, what is the image of $-sigma_1$? And that of $sigma_{1,1}$? I never did such a calculation..
$endgroup$
– Cla
Jul 17 '14 at 9:25
3
3
$begingroup$
The natural map $CH(G)rightarrow H^*(G,mathbb{Z})$ is a ring isomorphism.
$endgroup$
– abx
Jul 15 '14 at 14:28
$begingroup$
The natural map $CH(G)rightarrow H^*(G,mathbb{Z})$ is a ring isomorphism.
$endgroup$
– abx
Jul 15 '14 at 14:28
$begingroup$
I know that a such map exists, but I don't how it works. So in my specific example, what is the image of $-sigma_1$? And that of $sigma_{1,1}$? I never did such a calculation..
$endgroup$
– Cla
Jul 17 '14 at 9:25
$begingroup$
I know that a such map exists, but I don't how it works. So in my specific example, what is the image of $-sigma_1$? And that of $sigma_{1,1}$? I never did such a calculation..
$endgroup$
– Cla
Jul 17 '14 at 9:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The cohomology ring of $G(k,n)$ is isomorphic to
$$frac{mathbb{Z}[c_1(T),...,c_k(T),c_1(Q),..., c_{n−k} (Q)]}{(c(T)c(Q) = 1)},$$
where $T$ and $Q$ are respectively the tautological and the quotient bundle.
The chern classes of the tautological and the quotient bundle are given in terms of Schubert cycles by the following formulas:
- $c_i(T) = (−1)^isigma_{1,...,1}$,
- $c_i(Q) = sigma_i$.
answered Jul 15 '14 at 15:54
A. M. Flat_LineA. M. Flat_Line
37916
$endgroup$
1
$begingroup$
In the displayed formula, you should have a $mathbb{Z}$ where you have a "coefficient field" $k$.
$endgroup$
– Jason Starr
Jul 15 '14 at 17:15
$begingroup$
You are right. I corrected it.
$endgroup$
– A. M. Flat_Line
Jul 15 '14 at 19:16
add a comment |
Your Answer
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1 Answer
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oldest
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1 Answer
1
active
oldest
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oldest
votes
$begingroup$
The cohomology ring of $G(k,n)$ is isomorphic to
$$frac{mathbb{Z}[c_1(T),...,c_k(T),c_1(Q),..., c_{n−k} (Q)]}{(c(T)c(Q) = 1)},$$
where $T$ and $Q$ are respectively the tautological and the quotient bundle.
The chern classes of the tautological and the quotient bundle are given in terms of Schubert cycles by the following formulas:
- $c_i(T) = (−1)^isigma_{1,...,1}$,
- $c_i(Q) = sigma_i$.
answered Jul 15 '14 at 15:54
A. M. Flat_LineA. M. Flat_Line
37916
$endgroup$
1
$begingroup$
In the displayed formula, you should have a $mathbb{Z}$ where you have a "coefficient field" $k$.
$endgroup$
– Jason Starr
Jul 15 '14 at 17:15
$begingroup$
You are right. I corrected it.
$endgroup$
– A. M. Flat_Line
Jul 15 '14 at 19:16
add a comment |
$begingroup$
The cohomology ring of $G(k,n)$ is isomorphic to
$$frac{mathbb{Z}[c_1(T),...,c_k(T),c_1(Q),..., c_{n−k} (Q)]}{(c(T)c(Q) = 1)},$$
where $T$ and $Q$ are respectively the tautological and the quotient bundle.
The chern classes of the tautological and the quotient bundle are given in terms of Schubert cycles by the following formulas:
- $c_i(T) = (−1)^isigma_{1,...,1}$,
- $c_i(Q) = sigma_i$.
answered Jul 15 '14 at 15:54
A. M. Flat_LineA. M. Flat_Line
37916
$endgroup$
1
$begingroup$
In the displayed formula, you should have a $mathbb{Z}$ where you have a "coefficient field" $k$.
$endgroup$
– Jason Starr
Jul 15 '14 at 17:15
$begingroup$
You are right. I corrected it.
$endgroup$
– A. M. Flat_Line
Jul 15 '14 at 19:16
add a comment |
$begingroup$
The cohomology ring of $G(k,n)$ is isomorphic to
$$frac{mathbb{Z}[c_1(T),...,c_k(T),c_1(Q),..., c_{n−k} (Q)]}{(c(T)c(Q) = 1)},$$
where $T$ and $Q$ are respectively the tautological and the quotient bundle.
The chern classes of the tautological and the quotient bundle are given in terms of Schubert cycles by the following formulas:
- $c_i(T) = (−1)^isigma_{1,...,1}$,
- $c_i(Q) = sigma_i$.
answered Jul 15 '14 at 15:54
A. M. Flat_LineA. M. Flat_Line
37916
$endgroup$
The cohomology ring of $G(k,n)$ is isomorphic to
$$frac{mathbb{Z}[c_1(T),...,c_k(T),c_1(Q),..., c_{n−k} (Q)]}{(c(T)c(Q) = 1)},$$
where $T$ and $Q$ are respectively the tautological and the quotient bundle.
The chern classes of the tautological and the quotient bundle are given in terms of Schubert cycles by the following formulas:
- $c_i(T) = (−1)^isigma_{1,...,1}$,
- $c_i(Q) = sigma_i$.
answered Jul 15 '14 at 15:54
A. M. Flat_LineA. M. Flat_Line
37916
answered Jul 15 '14 at 15:54
A. M. Flat_LineA. M. Flat_Line
37916
answered Jul 15 '14 at 15:54
A. M. Flat_LineA. M. Flat_Line
37916
answered Jul 15 '14 at 15:54
A. M. Flat_LineA. M. Flat_Line
37916
37916
1
$begingroup$
In the displayed formula, you should have a $mathbb{Z}$ where you have a "coefficient field" $k$.
$endgroup$
– Jason Starr
Jul 15 '14 at 17:15
$begingroup$
You are right. I corrected it.
$endgroup$
– A. M. Flat_Line
Jul 15 '14 at 19:16
add a comment |
1
$begingroup$
In the displayed formula, you should have a $mathbb{Z}$ where you have a "coefficient field" $k$.
$endgroup$
– Jason Starr
Jul 15 '14 at 17:15
$begingroup$
You are right. I corrected it.
$endgroup$
– A. M. Flat_Line
Jul 15 '14 at 19:16
1
1
$begingroup$
In the displayed formula, you should have a $mathbb{Z}$ where you have a "coefficient field" $k$.
$endgroup$
– Jason Starr
Jul 15 '14 at 17:15
$begingroup$
In the displayed formula, you should have a $mathbb{Z}$ where you have a "coefficient field" $k$.
$endgroup$
– Jason Starr
Jul 15 '14 at 17:15
$begingroup$
You are right. I corrected it.
$endgroup$
– A. M. Flat_Line
Jul 15 '14 at 19:16
$begingroup$
You are right. I corrected it.
$endgroup$
– A. M. Flat_Line
Jul 15 '14 at 19:16
add a comment |
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3
$begingroup$
The natural map $CH(G)rightarrow H^*(G,mathbb{Z})$ is a ring isomorphism.
$endgroup$
– abx
Jul 15 '14 at 14:28
$begingroup$
I know that a such map exists, but I don't how it works. So in my specific example, what is the image of $-sigma_1$? And that of $sigma_{1,1}$? I never did such a calculation..
$endgroup$
– Cla
Jul 17 '14 at 9:25